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mathcalculus

find the critical points: s(t)= (t-1)^4 (t+5)^3

  • one year ago
  • one year ago

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  1. mathcalculus
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    first we find the derivative.. i know that.. then...?

    • one year ago
  2. Yahoo!
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    then equate it to 0

    • one year ago
  3. jennychan12
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    set it = 0

    • one year ago
  4. mathcalculus
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    but wait, im confused to find the derivative for this one..

    • one year ago
  5. Yahoo!
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    s'(t) =0

    • one year ago
  6. jennychan12
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    chain rule/product rule

    • one year ago
  7. mathcalculus
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    its 4(t-1)^3 3(t+5)^2

    • one year ago
  8. Yahoo!
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    Yup

    • one year ago
  9. mathcalculus
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    what's next after it? i'm a little lost after it

    • one year ago
  10. Yahoo!
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    4(t-1)^3 3(t+5)^2 = 0

    • one year ago
  11. Yahoo!
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    Now Find t

    • one year ago
  12. mathcalculus
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    what about chain rule?

    • one year ago
  13. jennychan12
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    doesnt apply here because derviative of t is just 1

    • one year ago
  14. Yahoo!
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    if ur function is of the type (t^2 + 2)^4 then s'(t) = 4 (t^2+2) * 2t

    • one year ago
  15. Yahoo!
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    sorry s'(t) = 4 (t^2+2)^3 * 2t

    • one year ago
  16. mathcalculus
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    oh right.

    • one year ago
  17. mathcalculus
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    so if the derivative of t is 1 then how do i solve t?

    • one year ago
  18. mathcalculus
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    sorry imm not sure what to do after ... 4(t-1)^3 3(t+5)^2

    • one year ago
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